Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Ask your questions and receive precise answers from experienced professionals across different disciplines. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

An ice cream shop sells cones at the volume of 94.2 cubic meters they want to double the volume of the cones without changing the diameter of the cone so that ice cream scoop will stay on the top of the cold what should the dimensions of the new cone be at the old cone had a height of 10 centimeters

Sagot :

The volume of a cone is [tex]V=\pi r^2\frac{h}3[/tex] where r = radius and h = height. If the cone has a volume of 94.2 cm³ (I assume you didn't mean m³ because that would be ridiculously huge) and a height of 10 cm, we can plug these values into the formula to find the radius. Don't do any rounding.

[tex]94.2 = \pi r^2\frac{10}3 \\ 282.6 = \pi r^2 *10 \\ 28.26 = \pi r^2 \\ 8.99543738355 = r^2 \\ 2.99923946752=r[/tex]

Now we know that's going to be the radius of our new cone as well since we're keeping the diameter the same. The volume is going to be double 94.2 which is 188.4. Let's solve for the height.

[tex]188.4 = \pi (2.99923946752)^2\frac{h}3 \\ 188.4 = \pi(8.99543738355)\frac{h}3 \\ 188.4 = 28.26\frac{h}3 \\ 565.2 = 28.26h \\\\ \boxed{h = 20, r\approx 3}[/tex]